Math 152 diary, spring 2007
In reverse order: the most recent material is first.

Wednesday, April 4 (Lecture #20)
I hurried through the end of the discussion of the Fibonacci numbers. Again, I regret that the restictions imposed by the pace of the course prevent me from showing details of really interesting and useful examples.

Google gives more than four million pages in response to the word fibonacci. The highest ranking page has a great deal of information. Fibonacci's original publication was in 1202. The web site does not seem to discuss how the numbers occurred a thousand years earlier in Sanskrit literature, where they were used in the analysis of how long and short vowel sounds can be combined to form phrases!

This concludes presentation of material which will be included on the exam next week. But perhaps the most important single result of the course will now be presented, Taylor's Theorem.

I know about a half-dozen ways to discuss this material. The method I used in class, and will discuss here, is very slick and sneaky. Its chief advantage is that it is very fast, but its chief disadvantage is that students may just be astounded. The discusssion is not well motivated (why the heck is he writing that thing?). Right now, I think the speed is useful. I will use the time "saved" to present many computations using Taylor's Theorem. So here we go.

Choose a and x, two numbers, and then define a constant K by the equation
So K is the number which makes that equation true. If a is not equal to x, then a-x is not equal to 0 and you can solve for K in the equation, so there will be exactly one K satisfying the equation. The object of what we're doing is to discover another description of K. Well, now define a function F(w) by the equation
and w is a variable between a and x. The colors in the pieces of the formula (if you can see them) are used so that the next step may be more understandable.
Differentiate this function F(w) with respect to the variable w. Be sure to differentiate everything that contains a w. The product rule must be used twice, and the Chain Rule gives some minus signs.
     F´(w)=f´(w)+f´´(w)(x-w)-f´(w)+ (f(3)(w)/2)(x-w)2-(f´´(w)/2)·2(x-w)2+K[3(x-w)2(-1)]/6
There are f´(w) and -f´(w) which cancel. And there are two more terms which also cancel: f´´(w)(x-w) and -(f´´(w)/2)·2(x-w)2 (since the two 2's cancel!). F´(w)$ is actually given by

Let's learn a bit more about F(w). If we substitute x for w, we get:
because of the various appearances of (x-x). Now consider F(a), where we substitute a for w in the formula for F:
But we chose K so that this would be f(x).

Therefore we have a function, F(w), so that F(a)=f(x) and F(x)=f(x). F(w) has the same value at both a and x. Let's use the Mean Value Theorem. It states that there is some c between a and x so that
Since F(x) and F(a) have the same value, then there is some c between a and x so that F´(c)=0. But F´(w) is described above so:
This is 0. We can solve for K now. Do it (multiply by 2, divide by (x-c)2, put the K on the other side of the equation. This gets us exactly what we wanted, another description of K. So K=f(3)(c) and now we know that
This is a version of Taylor's Theorem. Here is a more general statement.

Taylor's Theorem with Lagrange's form of the remainder
There is some number c between a and x so that f(x)=j=0n[f(j)(a)/j!](x-a)j + [f(n+1)(c)/(n+1)!]{(x-a)n+1.

This is neat and easy to remember because the remainder or error term looks almost like the next piece of what we are adding up. There is, of course, yet more vocabulary.

How to say it with infinite seriesThe Taylor language
(Here a is fixed and x is a variable and n is a positive integer.)
j=0n[f(j)(a)/j!](x-a)j is the partial sum of
j=0infinity[f(j)(a)/j!](x-a)j, a power series centered at a.
(Again, a is fixed, x is a variable, and n is a positive integer.)
j=0n[f(j)(a)/j!](x-a)j is the nth degree Taylor polynomial for f centered at a.
The series j=0infinity[f(j)(a)/j!](x-a)j is the Taylor series for f centered at a.
j=n+1infinity[f(j)(a)/j!](x-a)j is the infinite tail of the infinite series, and represents the difference between the sum of all the terms (a limit) and the sum of finitely many terms. [f(n+1)(c)/(n+1)!]{(x-a)n+1 (for some c between a and x) is the difference between f(x) and the value at x of the nth degree Taylor polynomial centered at a. It is called the remainder or error term.

How is a proof or idea discovered?
So how did the original ideas about Taylor's Theorem occur? What I just showed you is very contrived ("showing effects of planning or manipulation"). Certainly it seems to need knowledge of the final statement in order to prove the statement! I suspect that many specific examples were considered when people wanted to find "nice" polynomial approximations. I also suspect that reasoning using l'Hopital's rule occurred. I will discuss this next time and try to show how one could "guess" Taylor's Theorem, maybe.

I am not an expert. Brook Taylor lived in England from 1685 to 1731. Here is a link to a biography of Taylor, and here is one paragraph from that biography:

We must not give the impression that this result [Taylor's Theorem and Taylor series] was one which Taylor was the first to discover. James Gregory, Newton, Leibniz, Johann Bernoulli and de Moivre had all discovered variants of Taylor's Theorem. Gregory, for example, knew that
arctan x=x-(x3/3)+(x5/5)-(x77)+...
and his methods are discussed in [13]. The differences in Newton's ideas of Taylor series and those of Gregory are discussed in [15]. All of these mathematicians had made their discoveries independently, and Taylor's work was also independent of that of the others. The importance of Taylor's Theorem remained unrecognised until 1772 when Lagrange proclaimed it the basic principle of the differential calculus. The term "Taylor's series" seems to have used for the first time by Lhuilier in 1786.
I did not know until fairly recently that, five hundred years ago, there was detailed knowledge of Taylor series away from western Europe. A friend of mine, Professor David Bressoud of Macalester College in Minnesota, wrote an article whose title is Was Calculus Invented in India? (in the College Math Journal, 33 1, Pages 2-13, 2002). Here is the opening paragraph of his article:
No. Calculus was not invented in India. But two hundred years before Newton or Leibniz, Indian astronomers came very close to creating what we would call calculus. Sometime before 1500, they had advanced to the point where they could apply ideas from both integral and differential calculus to derive the infinite series expansions of the sine, cosine, and arctangent functions:
sin x=x-(x3/3!)+(x5/5!)-(x7/7!)+...
cos x=1-(x2/2!)+(x4/4!)-(x6/6!)+...
arctan x=x-(x3/3)+(x5/5)-(x77)+...
The traditional introduction of calculus is as a collection of algebraic techniques that solve essentially geometric problems: calculation of areas and construction of tangents. This was not the case in India. There, ideas of calculus were discovered as solutions to essentially algebraic problems: evaluating sums and interpolating tables of sines.

Geometry was well developed in pre-1500 India. As we will see, it played a role, but it was, at best, a bit player. The story of calculus in India shows us how calculus can emerge in the absence of the traditional geometric context. This story should also serve as a cautionary tale, for what did emerge was sterile. These mathematical discoveries led nowhere. Ultimately, they were forgotten, saved from oblivion only by modern scholars.

The article concludes with the following paragraphs.
There is no evidence that the Indian work on series was known beyond India, or even outside Kerala, until the nineteenth century. Gold and Pingree assert ... that by the time these series were rediscovered in Europe, they had, for all practical purposes, been lost to India. The expansions of the sine, cosine, and arc tangent had been passed down through several generations of disciples, but they remained sterile observations for which no one could find much use.

No. Calculus was not invented in India. Much of what we call calculus had been discovered, but the context for understanding these discoveries was never constructed. I am left wondering how much important mathematics today is known but not yet understood, passed among a coterie of tightly knit disciples as an intriguing yet seemingly useless insight, lacking the context, the fertilizing connections, that will enable it to blossom and produce its fruit.

I think it is useful to recognize that all sorts of human beings have intelligence and talent. Some study of history makes this clear.

Let me use Taylor's Theorem. So:

f(x)=j=0n[f(j)(a)/j!](x-a)j + [f(n+1)(c)/(n+1)!]{(x-a)n+1.
My candidate for a first "use" of Taylor's Theorem is sin(x). I like sine because its derivatives are really easy to compute. Just look. Also, I and almost all other folks use a=0. The values of the derivatives of sine at 0 are also displayed. Hey, just knowing four of these is enough to know all of these!
Now I wanted write the 11th degree Taylor polynomial for sine, centered at a=0. Here it is (notice the 0/1/0/-1 coefficients):
sin x=x-(x3/3!)+(x5/5!)-(x7/7!)+(x9/9!)-(x11/11!)
What about the error? This is (sin(12)(c)/12!)x12. I'll take the absolute value, and also notice that any derivative of sine has absolute value at most 1. So the error is at most x12/12!.

Let's see some numbers.

  • If x=1/2 (or .5 as some decimalists might write) then the 11th degree polynomial above has value 0.4792255386 and sin(.5) is (10 decimal place accuracy) 0.4794255386. Indeed. The error bound computed above, (.5)12/12!, is about 5·10-13.
  • If x=2 then the 11th degree polynomial above has value 0.909296136 and sin(.5) is (10 decimal place accuracy) 0.9092974268. Indeed. The error bound computed above, 212/12!, is about 8·10-6.

    The error overestimate is |x|n+1/(n+1)!. This depends on both x and n, and you can see the effect of the dependence in the computations above when n=11 and x=.5 and x=2. What is perhaps more interesting is the following, which I certainly hope you believe by now:
    If x is fixed, and n-->infinity, then |x|n+1/(n+1)!-->0.
    This is true because eventually n gets larger than x, and then the effect of multiplying by |x| on top compared to multiplying by stuff bigger than x on the bottom shows that the terms must approach 0. It may not be clear what the heck is going on. I promise that I will do many examples and applications. But here are three conclusions which I can write:

    The first result about where the series converges can be gotten by using the Ratio Test. I believe that any one who has attended class and done homework for the last two weeks would be able to verify the result. But why should any person be interested in this series? Or, perhaps more precisely, why should this series be more interesting than many other candidates? I don't know how, without using Taylor's Theorem, any way to tell you that this series is actually extremely interesting, and that its values, the sums of the series, are values of sine. Also, the third conclusion describes how rapidly and accurately the values of sine can be computed using partial sums. It shows that the approximating rate is actually very fast, and that these sums are quite practical for "real" computation. There's much more to come.

    Monday, April 2 (Lecture #19)
  • What is a power series? A power series centered at a is a series of the form
  • Where does it converge? Each series converges in an interval centered at a. This is called the interval of convergence. The distance from the center to the boundary of the interval is called the radius of convergence. The real behavior of the series is contained in the collection of coefficients, {cn}. The "a" just moves the center of the interval around.

    A collection of examples
    Power seriesCoefficientsInterval of convergenceRadius of convergence
    n=0infinityxn/nn cn=1/nn All real numbers: (-infinity,infinity) R=infinity
    One way to see this is to use the Root Test. Then the nth root of |xn/nn| is |x/n| and as n-->infinity, this-->0 for all x.
    n=0infinityxn/n2 cn=1/n2 [-1,1] R=1
    The Ratio Test results in considering |x|[n2/(n+1)2] and as n-->infinity, this-->1. So we get convergence for |x|<1. If x=+/-1, the series is a p-series with p=2>1, so it converges.
    n=0infinityxn/n cn=1/n [-1,1) R=1
    Use the Ratio Test again. The resulting ratio is |x|[n/(n+1)]. As n-->infinity, this-->1. So we get convergence for |x|<1. If x=1, divergence (Harmonic Series). If x=-1, convergence (Alternating Harmonic Series).
    n=0infinityx2n/n c2n=1/nn and codd=0 (-1,1) R=1
    The Ratio Test. Now when x=+/-1, both endpoints give the Harmonic Series, so the power series diverges at both endpoints.
    n=0infinitynnxn cn=nn Only {0} R=0
    Use the Root Test.

    It turns out that the qualitative behaviors displayed above are all that is possible for power series. That is, they must converge inside an interval, which can be all large as all of R or as small as the center of the power series. They may or may not, depending on the specific coefficients, converge or diverge at the end points of the interval. But no other types of behavior are possible.

    Powerful general facts

  • The sum is continuous.
    Suppose f(x)=n=0infinitycnxn. Inside the interval of convergence, f(x) is a continuous function.
    This is not too difficult to verify and can be done by looking at infinite tails and comparing the resutl to power series.
  • The sum is differentiable and the derivative is ...
    Suppose f(x)=n=0infinitycnxn. Inside the interval of convergence, f(x) is a differentiable function, and f´(x)=img src="gifstuff/sigma2.gif" width=13>n=0infinityncnxn-1.
    This is definitely more difficult to check.
  • And an antiderivative of the sum is ...
    Suppose f(x)=n=0infinitycnxn. Inside the interval of convergence, the series n=0infinitycnxn+1/(n+1) converges and its sum is an antiderivative of f(x).

    These results say that power series, inside their intervals of convergence, can be treated, for the purposes of calculus, just like "big" polynomials: differentiationk, integration, etc.

    "Reverse engineering" the result of the previous lecture
    Last time we needed to compute a "payoff" or expectation or fair entry fee for a gambling game: n=1infinityn/2n. How can we think about this? I'd like to show you how I "invented" what I did last time.

    Here is what I would do. I'd say to myself that this is a series which results from substituting x=1/2 into the power series n=1infinitynxn=x+2x2+3x3+4x4.... but this series is the result of multiplying the series 1+2x+3x2+4x3+... by x. So what is the sum of 1+2x+3x2+4x3+...=n=1infinitynxn-1? Well, nxn-1 is the derivative of xn, so the series 1+2x+3x2+4x3+... is the derivative of x+x2+x3+x4+... which is a geometric series with first term=x and ratio=x, so that its sum is x/(1-x).

    Now go backwards! The sum of the series n=1infinityn/2n is the result of differentiating x/(1-x), then multiplying the result by x, and then substituting in x=1/2. And that's what we did.

    And now another one ...
    A slightly more involved process would get the sum of n=1infinityn2/2n. Here look at the power series n=1infinityn2xn. Pull out an x to get n=1infinityn2xn-1 which is the derivative of n=1infinitynxn. But, hey, we just got the sum of that series (look at the previous paragraph). So we can get the sum of this series.

    Connecting the function and the coefficients
    If f(x)=n=0infinitycnxn=c0+c1x1+c2x2+c3x3+c4x4+... then:
    Set x=0 and get f(0)=c0.
    Differentiate the previous series. The result is f´(x)=n=0infinityncnxn-1=0+c1+2c2x+3c3x2+4c4x3+... then:
    Set x=0 and get f´(0)=c1.
    Differentiate the previous series. The result is f´´(x)=n=0infinityn(n-1)cnxn-2=0+0+2c2+6c3x+12c4x2+... then:
    Set x=0 and get f´´(0)=2c2 so that c2=f´´(0)/2.
    Differentiate the previous series. The result is f(3)(x)=n=0infinityn(n-1)(n-2)cnxn-3=0+0+0+6c3+24c4x+... then:
    Set x=0 and get f(3)(0)=6c3 so that c3=f(3)(0)/6.
    By now most people in the class recognized the pattern: cn=f(n)(0)/n!
    Please note that most people like this formula and therefore accept that 0! should be 1, and that f(0), the zeroth derivative of f, should be the original function. Then the formula, as stated, is certainly correct for n=0.
    This says several things. First, if a function has a power series expansion, then it has exactly one power series expansion, because the coefficients are given by that formula. This means that any (valid!) way we get the power series expansion, the result must be the correct answer. Any trick, any contrivance is good.
    How useful is this formula? For certain functions, it can be very useful. But, in general, if you give me a "random" function (say, a quotient or a composition) then computing high derivatives is tedious and difficult because the expressions for the derivatives begin to expand more and more ("expression swell"). So maybe for such functions the formula just given is not so useful.

    Some examples
    We do know one nice formula for a series and the sum of a series. That's:
    a/(1-r)=a+ar+ar2+ar3+..., the geometric series.
    If we can "rearrange" a function so that it fits the geometric series template, then maybe we can get a geometric series. Here are some examples.

    1. f(x)=3/(x+5). Actually, 3/(x+5)={3/5}/(1+{x/5}) (dividing the top and bottom by 5) and this is the same as {3/5}/(1-[-{x/5}]). Now I hope you can see the a and the r: a={3/5} and r=-{x/5}. So
    f(x)={3/5}-{3/5}{x/5}+{3/5}{x/5}2-{3/5}{x/5}3+...= n=0infinity(-1)n+1(3/5n+1)xn.

    2. (problem #8 in section 11.9) f(x)=x/(4x+1). Only a little bit of "rearrangement" is needed: x/(4x+1)=x/(1-{-4x}) so a=x and r=-4x.

    3. (problem #9 in section 11.9) f(x)=x/(9+x2). Here look at x/(9+x2)=(x/9)/(1+{x2/9})=(x/9)/(1-[-{x2/9}]) so that a=x/9 and r=-{x2/9}.

    A real example
    I began a discussion of the Fibonacci numbers, which occur almost everywhere.

    Wednesday, March 28 (Lecture #18)
    Exam coming ...
    The second exam of the semester will be given two weeks from today, on Wednesday, April 11. More information will be available soon on the topics to be covered in the exam. This information will appear linked to the course web page.

    How to actually compute values of functions is a serious question, and ideas concerning computation are the object of much pure mathematics and most of applied mathematics. It's generally agreed that polynomials are computable. Evaluating a polynomial involves multiplication and addition. Just to be clear, a polynomial is a function defined by a formula such as 22.45-9x2+101x5. It is a sum of monomials ("x to a positive integer power"), each possibly multiplied by some coefficient. The degree of a polynomial is the highest integer exponent where the coefficient is not zero. So the degree of the polynomial just written is 5.

    A polynomial solution to an initial value problem
    Since polynomials are easy to compute, we should try to use them in calculus as much as possible. In calculus we integrate and differentiate. Certainly we should try to solve Initial Value Problems (IVP's) for Ordinary Differential Equations (ODE's) with Initial Conditions (IC's). Here is a simple example:
          ODE: dy/dx=y
          IC: y(0)=1
    Let's suppose that y=A+Bx+Cx2+Dx3+Ex4+Fx+... is a solution. What can be said about the coefficients A, B, C, D, E, F, ...?
    So the IC, y(0)=1 says that when we insert x=0, the formula should give us the value 1. But all of the terms with x's in them are 0, so that 1=A.
    What does the differential equation tell us? The formula for y tells us that y´=0+B+Cx+2Dx+3Ex2+4Fx3+... so we can compare the two "presentations". Since two polynomials will be equal exactly when their coefficients agree, we learn that:
    A=B (the constant term), which (using the previous value of A) tells us that 1=B.
    B=2C (the coefficient of the linear or x term), which (using the previous value of B) tells us that 1/2=C.
    C=3D (the coefficient of the quadratic or x2 term, which (using the previous value of C) tells us that 1/6=C.
    D=4E (the coefficient of the cubic or x3 term, which (using the previous value of D) tells us that 1/24=D.
    I'll bet that the alert student could answer this question:

  • What is the coefficient of x10 in y? Answer: 1/10!.
    The following question is more dubious, because ...
  • What is the degree of the polynomial expression for y? The "dubious" part of the question is the assumption that the expression we got for y is actually a polynomial! It isn't, exactly, because there is no highest degree beyond which the coefficients are all 0. This is some other kind of creature.

    That is, I hope you (by now!) recognize the pattern of the coefficients, so that
    Right now I will define 0! to be 1, so that my notation will be easier to use. With the definition (or understanding!) y can be written as n=1infinity(1/n!)xn.

    We learned a while ago that solutions for Initial Value Problems should be unique. So we see clearly that
    This is certainly not clear. I sort of weaseled (?) a reason why the infinite sum should be considered a solution of the IVP. So maybe the infinite sum "is" such a solution. But there is no reason to suspect that it has to represent ex. In fact, everything I've done is actually correct, and the implied conclusions (that the series converges, and that the sum of the infinite series is ex, and that computing partial sums can be efficiently done and provides a great way to compute values of the exponential function) are all totally correct.

    Definition of power series
    A power series centered at a is a series of the form
    Here the numbers cn are the coefficients of the series, and the whole thing looks like some sort of infinitely long polynomial.

    An example
    Here is a power series centered at a=0:
    For which x's does this series converge? Well, we just return to the thinking mode of the previous lecture, with an=xn/(3n+1), and we try to think of a strategy to investigate convergence. The prominence of xn in the formula for an makes me want to use either the Ratio Test or the Root Test. I always feel more at home (?) with the Ratio Test (personal preference) so I'll try that.

     |an+1|   --------
              3(n+1)+1          |x|n+1(3n+1)          (3n+1)
    ------ = -------------- = --------------- = |x|· -------
                |x|n           {3(n+1)+1}|x|n         (3n+3)
     |an|      ------
    Now we need to have n-->infinity. The result is |x|, which is what we called L in the previous lecture. The Ratio Test gives this information:
        If |x|<1, the series converges absolutely and therefore converges. Such x's are in the interval -1<x<1.
        If |x|>1, the series diverges. Such x's are either in the interval (-infinity,-1) or in the interval (1,infinity).
        If |x|=1, the Ratio Test gives no information, and the series may converge or diverge. There is no reason for either conclusion.

    What happens if |x|=1? We need to use other methods. There are two x's which must be considered.

  • If x=+1, the series n=0infinityxn/(3n+1) becomes n=0infinity1/(3n+1). Here is one method to analyze this series: use the Limit Comparison Test, with the "other" series being the Harmonic Series, n=0infinity1/n. The limit will exist (depending on where you put the series, the result is either 3 or 1/3). Since the Harmonic Series diverges, this series must also diverge.
  • If x=+1, the series n=0infinityxn/(3n+1) becomes n=0infinity(-1)n/(3n+1). Here is one method to analyze this series: we can use the Alternating Series Test. This series does satisfy all three conditions (the terms alternate in sign, they decrease, and their limit is 0) so the series converges.

    To the right is a picture (yes, I work diligently to "translate" almost any information into pictures, so just ignore if you are not a picture person!) of the real line. I have tried to indicate how the line is divided into two collections of points. The red points (with D) are where this series diverges, and the green points (with C) are where this series converges. The behavior at the boundary of the intervals is sort of difficult to draw (!) but I hope the use of ] and [ and ( and ) help a bit.

    Another example ...
    Here is a power series centered at a=4:
    For which x's will this series converge?
    My hope was that students would make their own mental connections between the work needed to detect where this series converges and the work we just did. For example:

    The chunk of text on the left should change to the chunk of text on the right.
    ... Well, we just return to the thinking mode of the previous lecture, with an=xn/(3n+1), and we try to think of a strategy to investigate convergence. The prominence of xn in the formula for an makes me want to use either the Ratio Test or the Root Test. I always feel more at home (?) with the Ratio Test (personal preference) so I'll try that.
     |an+1|   --------
              3(n+1)+1          |x|n+1(3n+1)          (3n+1)
    ------ = -------------- = --------------- = |x|· -------
                |x|n           {3(n+1)+1}|x|n         (3n+3)
     |an|      ------
    Now we need to have n-->infinity. The result is |x| ...
    ... Well, we just return to the thinking mode of the previous lecture, with an=(x-4)n/(3n+1), and we try to think of a strategy to investigate convergence. The prominence of (x-4)n in the formula for an makes me want to use either the Ratio Test or the Root Test. I always feel more at home (?) with the Ratio Test (personal preference) so I'll try that.
     |an+1|   --------
              3(n+1)+1          |x-4|n+1(3n+1)          (3n+1)
    ------ = -------------- = --------------- = |x-4|· -------
               |x-4|n	   {3(n+1)+1}|x-4|n         (3n+3)
     |an|      ------
    Now we need to have n-->infinity. The result is |x-4| ...

    The logic should go totally in parallel, with x changing to x-4. Another way of seeing what's going on is to consider the series n=0infinity(x-4)n/(3n+1) and to ask, as I tried to do, what's the simplest (?) value of x to worry about? I claimed that this value of x is 4 because x=4 makes all of the (x-4)some power equal to 0. So I hope this these considerations make the following conclusion believable.

    The series converges if |x-4|<1, which is the same as -1<x-4<1, which is the same as 3<x<5. The series diverges if |x-4|>1, which is the same as x<3 or x>5. The series diverges for x=5 and converges for x=3.

    And maybe another ...
    How about n=0infinity(x+20)n/(3n+1)? I hoped that students would "see" the answer, which is shown graphically to the left. With more definiteness, I declare that this series converges if |x+20|<1, which is the same as -1<x+20<1, which is the same as -21<x<-19. The series diverges if |x+20|>1, which is the same as x<-21 or x>-19. The series diverges for x=-19 and converges for x=-21.

    Actually, the coefficients know it all
    The information about how a series n=0infinitycn(x-a)n converges is really contained in its collection of coefficients, the cn's. The "center" x=a just moves it around. Therefore here (and in almost all of the applications I know!) I will almost always consider series whose center is 0, n=0infinitycnxn, and not even worry about the x=a case. If we need to be concerned with a=500 or a=-sqrt(2), well, then, I will do the computations around x=0 and then just translate the results to make things work. The example is (more or less!) generally correct
    I'll try to illustrate with some general reasoning here why people like power series. I'll give an argument to show that convergence questions follow some straightforward logic. Suppose that I have a power series n=0infinitycnxn which converges at x1, shown on the real line to the right. What can I say about convergence at x2, also shown to the right. All that I know is that x2 is closer to the origin than x1. And I do not tell you anything more about the coefficients of the series. What follows is a very important and also slightly tricky bit of reasoning.

    I know n=0infinitycnx1n converges. Then (look at the first entry in the left-hand column of the table which began the previous lecture!) cnx1n-->0 because the individual terms in any series which converges must get smaller. But this means that there is some (maybe big!) number M so that |cnx1n|<M (if there were not such an M, some collection of the terms of the series would grow and grow, and they couldn't approach 0).

    I want to know if the series converges at x2. I'll consider absolute convergence. Maybe that's a stricter condition, but it is easier to work with. Look:
    n=0infinity|cn|·|x2|n=n=0infinity|cn| |x1|n(|x1|/|x2|)n<n=0infinityM(|x1|/|x2|)n.
    The first equality occurs because |x2|=|x1|·(|x2|/|x1|) and the second inequality occurs because we're overestimating by the M on each term. Please realize that these sorts of estimates are commonly used in real computations! But the series we ended up with is a geometric series with ratio equal to |x1|/|x2| and this is less than 1 and therefore it must converge. The Comparison Test then implies that the series we began with, the power series at x2, must also converge. The only relationship we needed between x1 and x2 to make this work is that the distance to 0 of x2 is less than the distance to 0 of x1. The x2 can be on the other side of the origin (because only absolute values appear in the discussion!). If you'd like to put the idea in terms of disease or contagion or something, then all of the x2's closer to the origin than x1 "catch" convergence from x1. Maybe the picture to the right sort of illustrates the situation. (The only peculiarity is that the "opposite" point, -x1, may not have convergence, but that's because a geometric series with ratio=1 does not converge, so the argument presented here does not allow any conclusion.)

    Whats happening and why?
    The argument just presented is really the central part of the proof that a power series has what is called an interval of convergence, and the center of the power series sits at the middle of the interval of convergence.

    Back to the example
    In the case of the example just presented, n=0infinityxn/(3n+1), the interval of convergence is -1<=x<1, with 0, the center of the power series, sitting in the middle of the interval.

    What sort of function is the sum?
    It turns out that the function which is sum of the power series is very nice inside the interval of convergence: it is continuous (no breaks or jumps) and it is differentiable (very smooth). But before this is discussed in detail, I wanted to show people ...

    How to gamble
    What I discussed is presented here.

    Monday, March 26 (Lecture #17)
    The lecturer was not enchanted with what he believed was useful to teach today: many discussions of series {con|di}vergence. This is because he couldn't think of any "interesting" way to present the material. He did believe that showing students repeated examples might be useful. Most of the entries in the table below were written on the board at the beginning of the lecture. The entries regarding the Ratio Test and the Root Test were written after a discussion about how to detect when a series resembles a geometric series and how to "compute" (in a limiting sense) the disguised ratio in such a series. The limit for e was recalled to deal with a few series when the root test was applied.

    Convergence of seriesFacts to know
    If an converges, then an-->0.
    So if the limit is not 0 or if the limit does not exist, the series must diverge.

    Series with positive terms
    Comparison Test; Integral Test; Limit Comparison Test.

    Alternating series
    If three conditions are true, then the series converges.
    If some of the conditions are not true, generally there's no conclusion.

    Absolute convergence implies convergence.

    Ratio Test
    Consider an, and try to compute limn-->infinity|an+1|/|an|=L.
    If the limit exists, then: L<1 implies absolute convergence; L>1 implies divergence; L=1 gives no information.

    Root Test
    Consider an, and try to compute limn-->infinity|an|1/n=L.
    If the limit exists, then: L<1 implies absolute convergence; L>1 implies divergence; L=1 gives no information.

    Sequence facts

        rn-->0 if |r|<1.

        a1/n-->1 if a>0.



    Series facts

    Geometric series
    0infinityarn=1/{1-r} if |r|<1. Diverges if |r|>=1 and a is not 0.
    1infinity1/np converges if p>1 and diverges if p<=1.


  • Students must be able to recognize and distinguish between geometric series and p-series!
  • Consider the Ratio Test applied to the harmonic series 1infinity1/n. Here an=1/n, so the ratio |an+1|/|an| is (1/{n+1})/(1/n) which is n/(n+1) and L=1. For the series 1infinity1/n2, the corresponding ratio is (1/{(n+1)2})/(1/n2)=[n/(n+1)]2. So here L=1 also. Since the harmonic series diverges and the other series (p-series, p=2>1) converges, the "L=1" can't give decisive information. By the way, the Root Test applied to both of these series also has L=1, so the "no information" conclusion is certainly valid there, also.
  • The converse of the first entry in the table under "Convergence of Series" is certainly false. A series can have its individual terms-->0 but it can still diverge. The harmonic series is an example.

    What was done
    I believe we discussed
    11.6: 6 (look also at 5), 8, 12, 23, 27 (look also at 25) and 11.7: 1, 6, 13, 15, ...
    We tried to determine if these series converge absolutely, converge conditionally, or diverge.

    The QotD was 11.7: 17: does 1infinity(-1)n21/n converge or diverge?

    A possible vocabulary word: retrograde, meaning "move backward in an orbit" (under certain circumstances, planets being observed seem to move backwards in their orbits) and then also "move in a direction contrary to the usual one".

    Wednesday, March 21 (Lecture #16)
    This week's vocabulary words
  • dichotomy "A division into two especially mutually exclusive or contradictory groups or entities."
  • manipulative (as a noun) "In teaching or learning arithmetic: a physical object which may be manipulated to help demonstrate a concept or practise an operation."

    We've been dealing with results which really depend on the terms of the series being positive. For example, the Comparision Test. Here's a version:

    Hypothesis Suppose 0<=aj<=bj.
    Conclusion If j=1infinitybj converges, then j=1infinityaj converges.
    If j=1infinityaj diverges, then j=1infinitybj diverges.
    I tried to be more poetic (?) in class. Somehow I said something about child and parent, but now when I think about the analogy, it seems rather silly. Perhaps this is true about many of the things I say in class.
    The enormous importance of the assumption of positiveness is easy to show. Look at the really silly example of these two sums:
         j=1infinity-13j compared to j=1infinity0. The sum of all 0's certainly converges. But, golly, -13-26-39-52-65... sure does diverge. There is no floor to the partial sums -- they just fall away to minus infinity.

    Another comparison result
    Please look in the text for some good examples using the Limit Comparison Test which can be useful. There are numerous textbook questions which you can try. Here is a statement:

    Hypothesis Suppose aj and bj are both always positive, and that the limj-->infinityaj/bj exists and is positive.
    Conclusion Either both j=1infinityaj and j=1infinitybj converge or
    both j=1infinityaj and j=1infinitybj diverge.
    The limit hypothesis states that, more or less, when j is very large, the terms aj and bj have the same size. To me this makes the conclusion believable, since convergence/divergence is really about whether the infinite tails are finite or not, and that depends about the sizes of the terms when j is very large. The limit comparison test serves as motivation for some further results we will see.

    Series with terms of varying signs
    So now we will complicate things a bit, and look at series whose signs vary. Let me start really easily but things will get more intricate rapidly. (Varying stop signs)

    This is just about the simplest example I could show. We got a formula for the jth term. We need the sign to alternate, and that will be given by (-1)something here. The sign will alternate if the "something here" is either j or j+1. The first term will be +1 and the second term will be -1 if we use j+1. So an explicit formula is aj=(-1)j+1. Next I asked about convergences of the series j=1infinity(-1)j+1. For this we must consider the sequence of partial sums.
          s1=1; s2=1-1=0; s3=1-1+1=1; s4=0, etc.
    It isn't too difficult to see that seven integer=0 and sodd integer=1. The partial sums flip back and forth. This is exactly the kind of behavior we did not get when we considered series with all positive terms. There the partial sums just traveled "up", to the right. Well, this particular infinite series does not converge, since the partial sums do not approach a unique limit.
          j=1infinity(-1)j+1 diverges.

    This is a more complicated series. I suggested that we try to "guess" a formula by first getting a formula for the sign, and then a formula for the absolute value. In this case, the sign is surely given by (-1)j+1, just as before. The magnitude (these are 1-dimensional vectors, after all!) or absolute value is 1+{1/j} (some students suggested {j+1}/j, which is also a good formula. So putting these together, aj=(-1)j(1+{1/j}). And now we looked at the {con|di}vergence of j=1infinity(-1)j+1(1+{1/j}).

    The partial sums are more complicated and more interesting.
          s1=2; s2=2-(1{1/2})-{1/2}=.5; s3=2-(1{1/2})+{1{1/3})=11/6=1.8333; s4=2-(1{1/2})+(1{1/3})-(1{1/4})={7/12}=.58333
    This is where I stopped computing in class, but, golly, I have a friend who could compute s17 either exactly ({4218925/2450448}) or approximately (1.72169). This is nearly silly. Richard Hamming, one of the twentieth century's greatest applied mathematicians, remarked that

    The purpose of computing is insight, not numbers.

    Let's try to get some insight. Look at the first four partial sums on the number line.

    From s1 to s2, we move left since the second term in the series is negative. From s2 to s3 we move right, because the third term in the series is positive. But notice that we don't get to s1. because the jump right has magnitude 1{1/3} and this is less than 1{1/2}the magnitude of the previous jump left.

    I hope you are willing to believe that what's described persists in general.

  • The even partial sums are increasing.
  • The odd partial sums are decreasing.
  • All of the even partial sums are less than all of the odd partial sums.

    Does this series converge? Students in both classes had interestingly varied opinions about this, and I will admit that I tried to encourage discussion. But the collection of partial sums does not approach a unique limit. Why? Well, the distance between any odd partial sum and any even partial sum will be at least 1, since the magnitude of the jth term is 1{1/j}, which is certainly >1. The partial sums can't get close.
          j=1infinity(-1)j+1(1+{1/j}) diverges.

    Here aj has sign (-1)j+1 again, and the absolute value or magnitude is 1/j. Does j=1infinity(-1)j+1(1/j) converge? The partial sums are more complicated and more interesting.
          s1=1; s2=1-(1/2)-1/2=.5; s3=1-(1/2)+{1/3)=5/6=.8333; s4=1-(1/2)+(1/3)-(1/4)=7/12=.58333
    Here's a picture of these partial sums. Things are a bit more crowded (that's good for convergence!) than in the previous picture.

    The previous three qualitative properties still hold. Since the signs alternate, the partial sums wiggle left and right. Since the absolute values decrease, the odd sums are less than the even sums, and all of the even sums are less than all of the odd sums. But now the distance between the odd and even sums-->0 since the magnitude of the terms is 1/j, and this-->0. So here is a rather subtle phenomenon:
          j=1infinity(-1)j+1(1/j) converges.

    The theorem on alternating series (Alternating Series Test)
    The following is the major result of section 11.5 of the text.

    Hypotheses Suppose that
  • The terms of a series alternate in sign (+ to - to + etc.).
  • The absolute value or magnitude of the terms decreases.
  • The limit of the absolute values of these terms is 0.
    Conclusion The series converges.
  • This is a very nice result, and is useful for some special series. The most famous example is the alternating harmonic series, j=1infinity(-1)j+1(1/j), which we just saw. There are other examples in section 11.5.
    Notice that the alternating harmonic series converges but the original harmonic series with the signs stripped off, j=1infinity(1/j), diverges. To me this is somewhat subtle.

    Finally, here is some "experimental evidence" which might help you believe that the alternating harmonic series converges.

    Some partial sums of the
    alternating harmonic series

    In fact, as several people guessed, the sum of the alternating harmonic series is ln(2). But the convergence is actually incredibly slow. The one millionth partial sum (which took almost 8 seconds for a moderately fast PC to compute) only has 5 accurate decimal digits. This is not the best and fastest way to compute things!

    But what if ...
    The sign distribution of terms in an infinite series could be more complicated. I suggested that we consider something like

    j=1infinity ----------------------------
    Here the sign distribution of the top of the fraction defining aj is quite complicated. The first 20 signs are here:
          -1, 1, 1, -1, -1, -1, 1, -1, 1, -1, 1, 1, -1, 1, 1, -1, -1, 1, 1, -1
    There's no nice pattern that I can see. Does this series converge?

    Please notice that with a few modifications, the corresponding question can be answered very easily. Look at:

    j=1infinity ----------------------------------
    Absolute values signs have been put around the cosine and sine functions. Now the series has all non-negative terms and we can use our comparison ideas. How big is the top? Since the values of both sine and cosince are in [-1,1], the top can't be any bigger than 9. The bottom is 2j. Therefore each term of this series is at most 9/2j. But this larger series is a geometric series with ratio 1/2<1 and therefore it must converge.

    Proof via manipulative
    There was a spectacular demonstration in class! It wa inspired by thinking about old-fashioned folding carpenter's rulers. If we have an infinite series j=1infinityaj, we could consider the associated series j=1infinity|aj|, where we have stripped off the signed of the terms, and are just adding up the magnitudes. This is sort of like an unfolded carpenter's rule, stretched out as long as possible. It may happen that the series of absolute values, a series of positive terms, may converge. So when "the ruler" is stretched out as long as possible, it has finite length. Well, if we then fold up the ruler, so some segments point left (negative) and some point right (positive) then the resulting length is also positive.

    The picture to the left as an attempt to show this statement and to duplicate the physical effect of what I displayed in class. The top has the segments stretched out as far as possible. The next picture shows some of the segments rotating, aimed backwards (negatively). The last segment shows in reg segments which are negative and in green the other segments, oriented postively. I hope this makes sense, and justifies the following:
    If j=1infinity|aj| converges, then j=1infinityaj must converge also (and, actually, |j=1infinityaj|<=j=1infinity|aj|).

    Proof via algebra
    There is a verification of these statements in the textbook, using algebra, on p.741 in section 11.6, if you would like to read it. Sigh.

    And conversely?
    Notice that the converse of the assertion about absolute values and series may not be correct. That is, a series may converge, and the series of absolute values of its terms may not. The simplest example, already verified, used the alternating harmonic series, divergent, and the harmonic series, convergent.

    A series j=1infinityaj which has j=1infinity|aj| converging is called absolutely convergent. Then the correct implication above is:

    If a series is absolutely convergent, then it is a convergent series.
    A series for which j=1infinityaj and j=1infinity|aj| diverges is called conditionally convergent. The alternating harmonic series is conditionally convergent.

    Another example
    Consider j=1infinity{sin(5j+8)}37/j5. I don't know very much about {sin(5j+8)}37 except that, for any j, this is a number inside the interval [-1,1]. Therefore j=1infinity|{sin(5j+8)}37/j5| has terms which are all smaller than j=1infinity1/j5 (a p-series with p=5>1, so convergent)> The comparison test asserts that j=1infinity|(sin(5j+8)37/j5| converges, and therefore j=1infinity{sin(5j+8)}37/j5 itself must be a convergent series.

    Given a series, take absolute values
    The result just stated is a very powerful and easily used method. If you "give" me a series with random signs, the first thing I will do is strip off or discrad the signs and try to decide if the series of positive (non-negative, really) terms converges.

    Many of the classical "tests" are about how the series resembles the geometric series. So we think about how aj might look like arj for j large, and make conclusions based on this. Let's look at an example.

    An example from the text
    Here is problem #3 from section 11.6. It asks if this series is absolutely convergent, conditionally convergent, or divergent:
    The text uses n as the index of summation here, which shouldn't be too distressing. The text starts the sum from n=0 rather than n=1, but, again, this shouldn't be too bad, since {con|di}vergence depends on the infinite tail, and the initial terms don't affect that at all. More subtly, if you inspect the formula for the term in the series, the text uses 0!, which I don't think has been previously mentioned. 0! means 1, and the reason for that understand will be given later.
    The answer It turns out that this series does converge (absolutely!) and, actually, its sum is e10. This is not obvious, and, again, the reasoning will be given later. Forget this premature statement of the conclusion and try to look at the sum, first using a "primitive" approach.

    Let's strip off the signs. Then |an|=(10)n/n!. For example, |a7|=107/7!=10,000,000/5040=1984 (approximately). This is large. But what if we consider how |a20| and |a21| are related? Then we need to investigate how (10)20/20! changes into (10)21/21!. The top multiplies by 10, of course, but the bottom changes by a factor of 21. The change from |a20| to |a21| is a multiplication by 10/21. And further thought should show that the next change involves multiplication by 10/22, and then 10/23, etc. All of the changes after n=20 use multiplication by numbers less than 1/2. Certainly the series converges (compare with a geometric series whose ratio is 1/2<1). There is a way to "mechanize" these comparisions and it is called the Ratio Test. The Ratio Test
    Suppose we are given an infinite series with nth term an and suppose that limn-->infinity|an+1/an| exists. If this limit is less than 1, then the original series converges absolutely and therefore must converge. If the limit is greater than 1, the original series diverges. If the limit happens to be equal to 1, the Ratio Test does not supply any information about convergence or divergence.

    Return to the example from the text
    Let's use the Ratio Test on the series n=0infinity(-10)n/n!. So we need to consider a limit. Here it is:

                  | (n+1)! |                 10n+1n!                 10
    limn-->infinity----------- =limn-->infinity---------=limn-->infinity-----= 0
                   |(-10)n|                 (n+1)!10n                n+1
                   |  n!  |
    Since this limit exists and is 0, and 0<1, the original series converges absolutely, and must therefore converge.
    I hope you see some resemblance to what is discussed here and technique for using geometric series for estimation of infinite tails which we considered last time. Certainly the Ratio Test is very efficient, but it does not give the numerical information which can be gotten using the more elaborate computations done then.


    I noted that all of the parentheses were important -- they could not be ignored in the simplification process. The answer should be 8/(n+1)3.

    Monday, March 19 (Lecture #15)
    This lecture is an example of how "theoretical" concepts can suddenly change into techniques giving real computational information. This is very striking. The theory begins, and then there will be enough numbers for everyone.

    Series with positive terms
    Today we will consider series whose terms are positive or, at worst, non-negative. (In the next lecture we'll discuss what happens if we allow different signs, but for now, only +'s.) What can we say in general about series with positive (or even just non-negative) terms? Well, the sequence of partial sums is increasing, since we're just adding more and more non-negative terms. What can happen? One thing is that the sequence of partial sums can tend to infinity (hey, this is what happens to the silly infinite series j=1infinity1: the sequence of partial sums is unbounded and the series diverges. Another thing that can happen to a positive series is that the sequence of partial sums can tend to a limit (a non-negative finite limit). This happens, for example, with positive geometric series with ratio less than 1. Then the sequence of partial sums is bounded and the series converges. This is a consequence of the fact that "Bounded monotonic sequences converge".

    This theoretical alternative is everything.

    Comparing series with non-negative terms
    Suppose we have two infinite series, j=1infinityaj and j=1infinitybj. We also will assume that both of the series have non-negative terms. This means that all of the ajj's and all of the bj's are >=0.

    Hypothesis Suppose we know that the terms of one series are always smaller than the terms of the other series. So, specifically, suppose we know that for all j, aj<=bj.
    Conclusion Suppose sn=j=1naj of the series with smaller terms, and tn=j=1nbj is the nth partial sum of the series with larger terms. Then sn<=tn.

    Comparing the series and inheriting {con|di}vergence

    Please note that the logic is important here. Divergence is "inherited" upwards and convergence is "inherited" downwards. Converse implications are false in general. (Example: The divergent series j=1infinity1 is bigger than the convergent series j=1infinity1/2j.
    Please read section 11.4 of the textbook for more details on comparison results.

    Easy examples
    Making the bottoms bigger in fractions shrinks the value. So the fraction 1/(56+7j+2j) is always less than 1/2j. I know that j=1infinity1/2j converges, and therefore the series j=1infinity(56+7j+2j) must also converge.

    Making the tops larger makes the value larger. Since I know that j=1infinity1/j, the harmonic series, diverges, I must know that j=1infinity[3+sin(j)]/j also diverges. This is because 3+sin(j)>=2 always.

    The prototypes
    The examples you need to know are these:
          Geometric series a+ar+ar2+ar3+ar4+... converges if |r|<1. When |r|>=1 (and a is not zero!) then the series diverges.
          p-series 1+1/2p+1/3p+1/4p+1/5p+1/6p+... converges if p>1 and diverges if p<=1.
    You should be able to do most of the problems in section 11.4 with these in mind. You should practice. I want to discuss some useful but more complicated applications.

    Bounds on tails with integrals
    Look at j=1infinity1/(arctan(j)+j3). This is a horrible series. I don't know much about arctan(j) when j is a positive integer except that the values are between 0 and Pi/2. Compare this series with j=1infinity1/j3 whose individual terms are each larger than the original series. The integral test tells me that this series converges since 1infinity(1/x3)dx is finite. (Its value is 1/2: please check this!).

    So I know that j=1infinity1/(arctan(j)+j3) converges. Well, that's fine but suppose I really need to know what the sum is? For example, suppose I want to know the sum to 3 decimal places (error less than 0.001). Here is a strategy. I write the infinite sum as a sum of two pieces, a partial sum and an infinite tail.
          j=1infinity1/(arctan(j)+j3)=j=1n1/(arctan(j)+j3)+j=n+1infinity1/(arctan(j)+j3)=sn+the nth "infinite tail"
    The infinite tail is j=n+1infinity1/(arctan(j)+j3). But each piece of that sum is less than j=n+1infinity1/j3. Look again at the picture here. So we know:
    Here's a computation of the improper integral:
    ninfinity(1/x3)dx= limA-->infinitynA(1/x3)dx= limA-->infinity-1/(2x2)]nA= limA-->infinity-1/(2A2)+1/(2n2)= 1/(2n2).
    So the error between the true sum and the nth partial sum is less than 1/(2n2). How can I get this less than 0.001=1/1,000? Let's take n to be, say, 25. Then 2(25)2=1,250. That works.

    Therefore the sum of the series j=1infinity1/(arctan(j)+j3) is the same as j=1251/(arctan(j)+j3) to 3 decimal digit accuracy. The following Maple command and its answer took less than a hundredth of a second:

    I think the sum is approximately 0.744.

    Bounds on tails with geometric series
    Let's try a different sort of series. I'll write some partial sums:

    1 + ------
         2.5     (2.5)2 
    1 + ------ + ------
          1!       2!
         2.5     (2.5)2   (2.5)3 
    1 + ------ + ------ + ------
          1!       2!       3! 
         2.5     (2.5)2   (2.5)3   (2.5)4 
    1 + ------ + ------ + ------ + ------
          1!       2!       3!       4!
         2.5     (2.5)2   (2.5)3   (2.5)4   (2.5)5 
    1 + ------ + ------ + ------ + ------ + ------
          1!       2!       3!       4!       5!
    I hope you get the idea. The infinite tail corresponding to this series is j=n+1infinity(2.5)j/j!.

    The sum of the terms up to and including the term with (2.5)5/5! is 11.6705.
    The sum of the terms up to and including the term with (2.5)10/10! is 12.1817.
    The sum of the terms up to and including the term with (2.5)15/15! is 12.1824.
    The sum of the terms up to and including the term with (2.5)20/20! is 12.1824.

    I bet this series converges, and that its sum is 12.18 (two decimal place accuracy). I'll verify this by looking at the tail j=11infinityaj. Here aj=(2.5)j/j! The first term is a11 which is 0.000597. I computed this -- the value is not "obvious" from the formula. This is not too interesting, because I need information about all of the terms, and there are infinitely many. Let me investigate how the terms are related to one another in the following way:

     aj+1       (j+1)!      (2.5)j+1j!     (2.5)j!     2.5
    ------ =  ---------- = ----------- = --------- = -----
      aj        (2.5)j     (2.5)j(j+1)!   (j+1)j!     j+1
    There's a whole bunch of algebraic things happening in this collection of equations. First, when we write the formulas for the terms, we get a compound fraction (a fraction of fractions) which then is converted (carefully!) into a simple fraction (just one division). Then the powers mostly drop out. Finally, we need the definition of factorial in order to see how (j+1)! relates to j!.

    If j is at least 11, then 2.5/(j+1) will be at most 2.5/12 and this is less than 0.209. Therefore we know that if j is at least 11, then aj+1/aj<0.209, so that aj+1<aj(0.209). We can replace each term by 0.209 multiplied by the term before it, and this is an overestimate.

    Now what? Look at the tail more closely:
    Here what I mean by "etc." is that we can keep pushing the subscripts "down" until they reach 11, each time "paying" by multiplying by another power of 0.209. And these are successive overestimates. So we finally end up with a geometric series whose first term is a11=0.000597, with ratio=0.209. This series has sum equal to 0.000597/(1-0.209) which is about 0.00076. Hey! This number is less than .001, and I bet that the true sum of the series therefore has leading decimal expansion 12.181 or 12.182.

    Another example
    Here we go:

    1 + ------
         9.8     (9.8)2 
    1 + ------ + ------
          1!       2!
         9.8     (9.8)2   (9.8)3 
    1 + ------ + ------ + ------
          1!       2!       3! 
         9.8     (9.8)2   (9.8)3   (9.8)4 
    1 + ------ + ------ + ------ + ------
          1!       2!       3!       4!
         9.8     (9.8)2   (9.8)3   (9.8)4   (9.8)5 
    1 + ------ + ------ + ------ + ------ + ------
          1!       2!       3!       4!       5!
    I hope you get the idea. The infinite tail corresponding to this series is j=n+1infinity(9.8)j/j!.

    The sum of the terms up to and including the term with (9.8)10/10! is 10965.326.
    The sum of the terms up to and including the term with (9.8)20/20! is 18011.195.
    The sum of the terms up to and including the term with (9.8)30/30! is 18033.744.
    The sum of the terms up to and including the term with (9.8)40/40! is 18033.744.

    I bet this series converges, and that its sum is 18033.74 (two decimal place accuracy). I'll verify this by looking at the tail j=31infinityaj. Here aj=(9.8)j/j! The first term is a31 which is 0.000650. I computed this -- the value is not "obvious" from the formula. This is not too interesting, because I need information about all of the terms, and there are infinitely many. Let me investigate how the terms are related to one another in the following way:

     aj+1       (j+1)!      (9.8)j+1j!     (9.8)j!     9.8
    ------ =  ---------- = ----------- = --------- = -----
      aj        (9.8)j     (9.8)j(j+1)!   (j+1)j!     j+1
    There's a whole bunch of algebraic things happening in this collection of equations.First, when we write the formulas for the terms, we get a compound fraction (a fraction of fractions) which then is converted (carefully!) into a simple fraction (just one division). Then the powers mostly drop out. Finally, we need the definition of factorial in order to see how (j+1)! relates to j!.

    If j is at least 31, then 9.8/(j+1) will be at most 9.8/32 and this is 0.306 (exactly! wow!). Therefore we know that if j is at least 31, then aj+1/aj<0.306, so that aj+1<aj(0.306). We can replace each term by 0.306 multiplied by the term before it, and this is an overestimate.

    Now what? Look at the tail more closely:
    Here what I mean by "etc." is that we can keep pushing the subscripts "down" until they reach 31, each time "paying" by multiplying by another power of 0.306. And these are successive overestimates. So we finally end up with a geometric series whose first term is a31=0.000650, with ratio=0.306. This series has sum equal to 0.000650/(1-0.306) which is about 0.000936. Hey! This number is less than .001, and I bet that the true sum of the series therefore has leading decimal expansion 18033.744 or 18033.745.

    Random (?) facts
    e2.5 is approximately 12.18249396
    e9.8 is approximately 18033.74493

    Not at all random, and this will be explained later.

    The series j=1infinity1/(j5+3j) converges. I told students that j=1101/(j5+3j)=0.27951. Then I asked that students find an error estimate. So some sort of overestimate of j=11infinity1/(j5+3j) is needed. Several strategies can be used to get valid answers. Here are two strategies.

    Compare to a p-series and then estimate with an integral
    Certainly for any positive integer j, 1/(j5+3j)<1/(j5). Therefore j=11infinity1/(j5+3j)<j=11infinity1/(j5). This sum is, in turn, less than the improper integral 10infinity(1/x5)dx. Computation of the integral:
    10infinity(1/x5)dx=limA-->infinity10A(1/x5)dx=limA-->infinity-1/(4x4)10A=limA-->infinity-1/(4A2)+1/(4·104)=1/(4·104). This is a fine answer just as it is!

    Compare to a geometric series and then get the sum of the geometric series
    Certainly for any positive integer j, 1/(j5+3j)<1/3j. Therefore j=11infinity1/(j5+3j)<j=11infinity1/3j. But this infinite series is a geometric series whose first term is 1/311 and whose ratio is 1/3. The sum of this series is {1/311}/(1-{1/3}). This is a fine answer just as it is!

    Remarks about the QotD solutions

    1. Why should the improper integral have its lower bound equal to 10 rather than 11? A number of students used 11 as the lower bound. The aim is to overestimate the infinite tail j=11infinity1/(j5+3j) by something easier to compute. The most important word segment here is over in "overestimate". With this in mind, please look at the picture to the right. It should convince you that if you want to overestimate j=11infinity(1/j5) by something like ?infinity(1/x5)dx that the correct choice for ? is 10.
    2. Which error estimate is "better"? Except for rather rare circumstances, in real life getting any useful estimate is good enough. In this case, the first error estimate, 1/(4·104), is 0.000025, and the second error estimate, {1/311}/(1-{1/3}), is 0.00000847. I guess I would generally think that 1/3j will give something smaller than 1/j5, since exponentials grow faster than polynomials. But both estimates show that the number given for the tenth partial sum, 0.27951, is accurate to at least 4 decimal places.

    Wednesday, March 7 (Lecture #14)
    I briefly discussed some questions involving academic integrity. You may read about the Rutgers academic integrity policy.

    Again, a series is a formal sum, a1+a2+a3+... Your text writes this using sigmas, j=1infinityaj.

    One comment Please realize that the j doesn't really matter. It is what is called a dummy variable. The j is a logical placeholder. The sum k=1infinityak represents the same thing, and the sum q=1infinityaq represents the same thing, and so, I suppose, the sum which follows represents the same thing: TOAD=1infinityaTOAD (why can't we use words like TOAD as index variables?).

    Associated with each series are two different sequences. One sequence is the sequence of terms of the series: {aj}. Another sequence is the sequence of partial sums of the series: this sequence looks like this: a1, a1+a2, a1+a2+a3, a1+a2+a3+a4. So the 400th partial sum would be j=1400aj.

    A specific example: the harmonic series
    The harmonic series 1+1/2+1/3+1/4+1/5+...=j=1infinity1/j has these two series associated with it:

    The sequence of individual terms
    The first five terms: 1, 1/2, 1/3, 1/4, 1/5, .... I hope that the asymptotic behavior of this sequence whose jth term is 1/j as j gets large is obvious: the sequence of individual terms -->0.

    The sequence of partial sums
    The first five terms: 1, 1+1/2=3/2 (1.5), 1+1/2+1/3=11/6 (about 1.8333), 1+1/2+1/3+1/4=25/12 (about 2.0833), 1+1/2+1/3+1/4+1/5=137/60 (about 2.2833). The asymptotic behavior of this sequence is certainly not obvious or clear or ... We used some quite tricky reasoning last time to see that this sequence of partial sums is not bounded. They get really really large. I'll try to show this again later in today's lecture using a different approach which is maybe more systematic.

    A series converges if ...
    A series converges if the sequence of partial sums converges. The limit of the sequence of partial sums is frequently called the sum of the series.

    It can be difficult to decide if a series converges (heck, the harmonic series does not, even though the terms go to 0). Even if you know a series converges, it can be difficult to approximate the sum of the series. This is interesting, because most of the computations that are commonly done using calculators and computers (such as sin(.567) or e3.45) consist of numerical approximations to sums of infinite series. So what we are doing is interesting and useful.

    The infinite tail
    An infinite series j=1infinityaj can be thought of as j=1naj+j=n+1infinityaj. So there is the nth partial sum plus the "other" terms of the series. I like to think of this maybe as some sort of animal. The partial sum is the body, and the infinite tail is ... well, the tail. The question of whether the series converges or not maybe is analogous to whether the weight of the animal is finite (this is a good analogy only with series whose terms are all positive -- we will deal later with series whose terms change sign). The weight will be finite exactly when the infinite tails-->0 as n-->infinity. In fact, the first "few" terms of a series have nothing to do with convergence!

    For example, we already know that the harminic series 1+1/2+1/3+1/4+1/5+... does not converge. Then the series 56+37.8+409+1/4+1/5+1/6+... (all other terms are the terms in the harmonic series, unchanged) also does not converge. The sequence of partial sums has some change (1-->56, 1/2-->37.8, 1/3-->409) but the fact that the partial sums eventually do not approach a one fixed finite number is still correct.

    In a few minutes (geometric series) we will see that the series 1/2+1/4+1/8+1/16+... (here the nth term is 1/2n) does converge, and its sum is 1. If I change the first term to 15 and the third term to 88, the series still will converge. Its sum will be 1+(15-1)+(88-1/8) (the changes are made to the old sum).

    So really when we study convergence we are looking at infinite tails.

    Geometric series
    One kind of series has nice, simple formulas for partial sums and for sums, and that's geometric series. A geometric series consists of some sort of starting term, and each further term is formed by multiplying the previous term by some constant factor. So:

    1. A series
      5+5/7+5/72+5/73+5/74+... is a geometric series. The starting term is 5, and the constant factor, the multiplier, is 1/7. The use of "..." here is traditional. It is supposed to be telling the reader that, "Hey, I've given you enough evidence, enough of the pattern, you should now be able to figure out what the general formula of the series is." Sometimes the ... implied pattern is clear, and sometimes, darn it!, it is not.
    2. Another series
      3-3/19+3/(19)2-3/(19)3+3/(19)4-3/(19)4+3/(19)5... This is also a geometric series. Here the first term is 3, and the multiplier, usually called the ratio, between successive terms is -1/19.
    3. And another
      Look carefully, please, at 7+2·7/9+3·7/92+4·7/93+5·7/94+6·7/95+... and try to tell if this is a geometric series. In fact, looking at any three consecutive terms is enough to disqualify this series. For example, we start with 7 and the next term is 2·7/9. These terms dictate that the ratio is 2/9. But the term after 2·7/9 is 3·7/92 and these two terms indicate that the ratio must be (3/2)/9. Now 2/9 and 3/18 are not the same. The ratio between successive terms must be constant in a geometric series, so this series is not a geometric series.
    Partial sums and sums of geometric series
    Look at a+ar+ar2+ar3+...arn-1. I'll call this sn, the nth partial sum of the geometric series. We can do some arithmetic with sn. Look:
    Multiply by r:
    Almost all of the terms are the same, except for the first term of sn and the last term of rsn. So we can subtract, and many things cancel:
    But sn-rsn=(1-r)sn so that

    Formula for the nth partial
    sum of a geometric series
    sn= --------

    This is a neat formula which is sometimes very useful. Please realize that an explicit formula for the partial sum of a series is very, very rare. It should not be expected.
    We saw last time (look in section 11.1, page 707) that if |r|<1, then rn-->0 as n-->infinity. Therefore:

    Formula for the sum of a
    geometric series when |r|<1
    s = -------

    Use #1 of geometric series
    Here is something which many people are supposed to see (!) before college. The infinite repeating decimal 0.731731731... represents a rational number (a quotient of integers). What rational number does it represent?

    Here the most interesting problem is recognizing the implied geometric series. Decimal notation is very clever, and conceals some true subtleties. So 0.731 itself means 731·(.001) which is 731/1000. What about 0.000731? This is 731·(.000001) which is 731·(.001)·(.001). That is, 10-3·10-3=10-6. Therefore 0.000731 is 731/(1000)2. And similarly 0.000000731 is 731/(1000)3. So we see (maybe not so "clearly"!) that:
    We therefore recognize that the repeating decimal indicates an infinite series whose first term, a=731/1000, and whose constant ratio between successive terms is r=1/1000. The sum is then a/(1-r)=[731/1000]/(1-[1/1000])=[731/1000]/[999/1000]=731/1000.

    Digression: how maybe this is done in earlier "grades"
    A teaching might say the following:
    Let's consider the number Q=0.731731731... and try to figure out another way of looking at Q. Well, 1,000Q=(1,000)·(0.731731731...)=731.731731731... so then:
    1,000Q=731.731731731... and subtract
    The result is 999Q=731, so that Q=731/999. Ain't that nice! (Thanks to Ms. Panova for pointing out this approach.) My "excuse" for pointing out the geometric series approach is that I want to show you a use of geometric series, and also to maybe expose a bit of the structure of the decimal system, which is actually a very clever intellectual construction.

    Use #2 of geometric series
    A square of side length 5 has another square whose side length is half of that, placed outside but so that corners and an edge coincide. Another square whose side length is half of that, placed outside of both squares but so that corners and an edge coincide. And ...
    My language is perhaps not too precise. A sort of picture of this object (just the first 6 squares) is shown to the right. The object is an example of a fractal. General information about fractals is here and a source which is very accessible is here

    The question What is the total area of all of the squares?

    The first square has area 5·5=52. The second square has area (5/2)·(5/2)=52/4. The third square has area (5/2/2)·(5/2/2)=(5/22)·(5/22)=52/42

    The pattern may convince you that the total area is the sum of
    This is a geometric series whose first term is a=52, and the constant ratio between successive terms is r=1/4. The sum is then a/(1-r)=52/(1-[1/4])=100/3. This is the total area.

    I remarked to students that questions about some other geometric quantities can easily be asked. For example,
    The question What is the total perimeter of all of the squares?

    The first square has perimeter 4·5=20. The second square has perimeter 4·(5/2)=20/2. The third square has perimeter 4·(5/2/2)=4·(5/22)=20/22

    The pattern may convince you that the total perimeter is the sum of
    This is a geometric series whose first term is a=20, and the constant ratio between successive terms is r=1/2. The sum is then a/(1-r)=20/(1-[1/2])=40. This is the total perimeter.

    And here's another question. We could think about the region inside all of the squares as one part of the plane. Then the boundary between this region and the remainder of the plane could be analyzed. It has a border which is one relatively long horizontal line segment on the bottom, and one vertical line segment on the left. The region is shown to the right. I hope you "see" that it sort of fades off in a complicated way to the right. There's lots of border turnings on the right.

    The question What is the total length of the border of this region?

    One way to analyze this is to take the total perimeter of the squares, 40, which we got above, and to subtract the interior border lengths. Look at these pictures.

    The first inner wall has length 5/2. The second inner wall has length (5/2)/2=5/22. The third inner wall has length (5/22)/2=5/23.

    The pattern may convince you that the total length of the inner walls is the sum of
    This is a geometric series whose first term is a=(5/2), and the constant ratio between successive terms is r=1/2. The sum is then a/(1-r)=(5/2)/(1-[1/2])=5. This is the total length of the inner walls, so the perimeter of the region is 40-5=35. Whew! I think that's enough for this.

    Use #3 of geometric series
    Bruno and Igor have a loaf of bread. Bruno eats half the loaf and passes what remains to Igor. Igor eats half of what he is given and passes what remains to Bruno. Brundo eats half of what he is given and passes what remains to Igor. Igor eats half of what he is given and passes what remains to Bruno...

    The question How much bread (the total amount) does Bruno eat?

    As I mentioned in class, I know people who can somehow "solve" these problems by inspection, that is, they read or listen to the problem, and ZAP!!! the answer is clear. (The same is true for the geometric problems mentioned in Use #2.) I am not one of these "zap" people -- some of the students seem to be! I would probably solve the problem by computing the amount of bread Bruno and Igor eat, for at least a few rounds. I would try to discover the pattern, and then I'd use the observed pattern.
    Round #Bruno eatsIgor eats

    I filled out this table dynamically in class, with explanations being given as I did it. For example, I remarked that after Bruno ate half the loaf, Igor would receive the other half loaf. Igor would eat half of that, which is 1/4 loaf, and pass 1/4 loaf to Bruno. Bruno would eat half of a 1/4, which is 1/8 loaf, and pass the remaining 1/8 loaf to Igor, etc. It seems apparent ("clear") that Bruno eats 1/2+1/8+1/32+..., a quantity which we recognize as a geometric series. The first term, a, is 1/2, and the constant ratio between successive terms is 1/4. Therefore Bruno must eat a/(1-r)=(1/2)/(1-[1/4])=2/3 of the loaf. Poor Igor will eat 1-2/3=1/3 of the loaf.

    It is easy to change this problem. You could imagine the named people eating different quantities, or you could imagine there being more people, etc. Sums like this do arise in real applications, and I hope that you will be able to recognize them and cope with them.

    Integrals and sums
    Section 11.3 contains a full presentation of what I'll discuss here. I want to study the harmonic series again, and this time really convince you that it diverges. After that I want to study another series, convince you it converges, and then actually get a fairly good decimal approximation for its sum.

    The harmonic series again
    Let's look at 1+1/2+1/3+1/4+1/5+... but accompany this with a picture. In fact, let me begin with a specific partial sum, the fifth partial sum: s5= 1+1/2+1/3+1/4+1/5 (look, there are no ...'s!). I've sketched part of the graph of y=1/x to the right. I only sketched the graph on an interval beginning a bit less than 1 and ending a bit more than 6. The function f(x)=1/x is strictly decreasing. If I draw the boxes shown (this is the Riemann sum approximation to the area under the curve from x=1 to x=6 using a partition equally spaced with 5 subintervals and sample points at the left-hand endpoints, it just happens (NO! This is arranged very precisely.) that the Riemann sum approximation is equal to the fifth partial sum: the boxes all have widths equal to 1 and they have heights 1, 1/2, 1/3, 1/4, and 1/5. Therefore 161/x dx<s5. But I "know" the integral. It is ln(x)]16=ln(6) since ln(1)=0. Therefore ln(5)<s5.

    If you are curious, ln(6) is approximately 1.791 and s5 is approximately 2.28333. So the arithmetic reinforces the picture.

    Now I would like you to imagine a similar picture for y=1/x, where the interval goes from slightly to the left of 1 up to slightly to the right of n+1. Again, think of boxes over the graph, indicating a Riemann sum with left-hand endpoints at the integers, a bunch of boxes with width equal to 1. The function is still strinctly decreasing, so the tops of the boxes are above the graph. The sum of the areas is sn=j=1n1/j, the nth partial sum of the harmonic series. Therefore sn>1n+11/x dx=ln(n+1). But as n-->infinity, ln(n+1)-->infinity, and since sn is bigger than ln(n+1), sn-->infinity also, and the harmonic series diverges.

    I asked in class how big n should be for the nth partial sum of the harmonic series to be larger than, say, 200. I can be sure that sn>200 if I know that ln(n+1)>200 since sn>ln(n+1). But ln(n+1)>200 happens when (exponentiate both sides of the inequality) n+1>e200. How big is e200? It is about 7·1086. If you could do 1010 additions each second, then (there are 31,556,926 seconds in a year) to compute this partial sum would take about 2·1069 years. One estimate for the age of the universe is 13.7 billion years, or 1.37·1010 years. You'd need a lot of universes. I think it is amazing we can estimate such a partial sum.

    A convergent series
    Let's look at 1+1/24+1/34+1/44+1/54+.... It will turn out that this series converges. Let me first convince you of this with a geometric argument whose computations are based in calculus. So let me look at the graph of y=1/x4 on an interval from slightly to the left of 1 to slightly to the right of 6. Now let me consider Riemann sums with right-hand endpoints. Again each of the widths is 1 and the heights come from f(x)=1/x4 at x=2, 3, 4, 5, and 6. What we have (again, f(x) is strictly decreasing) is 161/x4dx>1/24+1/34+1/44+1/54+1/64. Again, let's compute the integral:
    The sum 1/24+1/34+1/44+1/54+1/64 is 0.08112. This isn't exactly s6 for this series, since we left out the first term, which is 1. In fact, though,

    A (possible) complaint!?
    This time we put the boxes under the curve. Before we put the boxes on top of the curve.
    How do we know where to put the boxes?
    And before we got a simple relationship between the integral and the partial sum, and now we've got to add the first term to the integral to get information connecting the integral and the partial sum.
    What's going on?
    If you are just exploring, sometimes you will need to look at the improper integral needed and see if it converges or diverges. You'll then need to adjust your logic to the situation. You can make mistakes, but these mistakes can be fixed. The situation really is very forgiving.

    To overestimate an infinite tail with an integral, put the boxes underneath the curve.
    To underestimate a partial sum with an integral, put the boxes on top of the curve.

    Now imagine the situation from 1 to n+1 with the graph of f(x)=1/x4. Similar consideration of boxes under the graph leads to
    Now 1n+11/x4dx=-1/(3x3)]1n+1=-1/(3(n+1)3+1/3
    As n-->infinity, this-->1/3. Therefore all of the sn's are bounded above by 1+(1/3). Since the sn's form an increasing sequence (hey, everything we're adding is positive!) we know ("Bounded monotonic sequences converge") that the sequence of partial sums {sn} converges, and therefore the infinite series converges!

    Yeah, but what is the sum?
    It happens that with more advanced methods the sum j=1infinity1/j4 can be computed exactly. You won't believe me but the value is Pi4/90, approximately 1.0823! The sums of most series can't be computed exactly in terms of "standard" constants (for example, if I changed the 4 to a 3 the series will still converge, and no one knows the exact value of the sum). Let me show you how to estimate an infinite tail.

    The partial sum j=1n1/j4 has an infinite tail beginning with the term 1/(n+1)4 and going on and on from there. If you consider the picture to the right, the infinite tail is overestimated by the improper integral ninfinity1/x4dx=(I will omit the details!)=1/(3n3).
    The infinite tail is the error between the partial sum and the true sum. If I want the partial sum to be accurate to 3 decimal places, we just need the infinite tail to be less than .001. I should find an n so that 1/(3n3) be less than 1/1,000. This will certainly happen when n=3. Therefore the true value of the sum will be approximated to 3 decimal places by s10. My silicon friend can compute this easily, and the value is 1.082.

    The integral test and p-series
    The textbook discusses the integral test in section 11.3. Here is a version of the integral test.

    The integral test
    Suppose f(x) is a positive decreasing function, defined for x>=1. Then the series j=1infinityf(j) converges exactly when the improper integral 1infinityf(x) dx converges.

    The logic behind the integral test is used to underestimate partial sums and to overestimate infinite tails, as previously discussed. The best-known application of the Integral Test is the following fact:

    Consider the infinite series
    where p is a constant. This series converges if p>1 and diverges otherwise.

    When p=1 this is the harmonic series which diverges. We analyzed the series for p=4 and showed that it converged.

    Monday, March 5 (Lecture #13)
    Examples of sequences
    Definition of the sequence First five elements of the sequence Limits and behavior
    an=1/n 1., 0.5000000000, 0.3333333333, 0.2500000000, 0.2000000000
    an=(-1)n -1., 1., -1., 1., -1.
    an=(1/2)n 0.5000000000, 0.2500000000, 0.1250000000, 0.06250000000, 0.0312500000
    an=10n 10., 100., 1000., 10000., 100000.
    an=[-1/2]n -0.5000000000, 0.2500000000, -0.1250000000, 0.06250000000, -0.03125000
    an=51/n 5., 2.236067977, 1.709975947, 1.495348781, 1.379729661
    an=n1/n 1., 1.414213562, 1.442249570, 1.414213562, 1.379729661
    an=n! 1., 2., 6., 24., 120.
    an=(50)n/n! 50., 1250., 20833.33333, 260416.6667, 2604166.667
    The sequence in Problem #64, section 11.1
    2., 1., 0.5000000000, 0.4000000000, 0.3846153846

    Big ideas

    Convergence and limit




    Monotonic & bounded implies ...


    The sequence of partial sums ...

    The harmonic series
    The partial sums 1., 1.500000000, 1.833333333, 2.083333333, 2.283333333 Primitive idea #1 It diverges because we're adding up infinitely many numbers, and therefore things get to be too darn large.
    Primitive idea #2 It converges because, although we are adding up lots of numbers, the steps between them get smaller and smaller and smaller, and so the sum can't get very large.

    So what does happen?

    Wednesday, February 28 (Lecture #12)

    There will be a quiz tomorrow. It will cover improper integrals and differential equations. Exponential decay
    dy/dt=ry y=Aert (here r is negative!) radioactive tracers in blood radiocarbon dating

    Exponential growth dy/dt=ry y=Aert (here r is positive!)

    QotD Bacteria

    Exponential growth with a "carrying capacity" Consider the differential equation dy/dt=y(2-y). If y is close to 0 but positive, then ... If y is close to 2 but less than 2, then ... Example of the Logistic differential equation, numbers selected to make everything easy. Let me try to solve this equation with the initial condition y(0)=1. Separate and integrate. The result (after use of partial fractions!) is -(1/2)ln(2-y)+(1/2)ln(y)=x+C Here C=0 (wow!). And then ln(y/[2-y])=2x, so y/[2-y]=e2x and then y=(e2x)[2-y]=2e2x-e2xy, so (1+e2x)y=2e2x and, finally, y=[2e2x]/[1+e2x]. This may be in a form that "real people" can understand.
    A picture of this curve is shown to the right. For large x, For x negative,

    Direction fields, a useful qualitative tool

    This is the direction field This is the direction field with
    some solution curves.
    This is the direction field This is the direction field with
    some solution curves.
    This is the direction field This is the direction field with
    some solution curves.
    This is the direction field This is the direction field with
    some solution curves.
    This is the direction field This is the direction field with
    some solution curves.

    Monday, February 26 (Lecture #11)
    Exams were returned today. One version of the exam is available, along with answers. Also available is information about the grading, including statistics about the grades, and brief remarks which students should read carefully about further work in the course.

    The 20% of the students who were not in class can come to my office before or after class to pick up your exam. They also can come to my office during an office hour to pick up your exam. Finally, they can come to my office at any other mutually convenient time to pick up your exam. (I did spend more than 12 hours this weekend grading the exams, in order to get them back to you in a timely manner.)

    Differential equations

    Differential equations, more of a definition

    Sample scenarios
    First story Salt dissolves in water. Put more salt in the water. It will dissolve, although perhaps less rapidly. Add more and more salt. Eventually the solution becomes saturated, and the salt just drops to the bottom (precipitates?). Differential equations can model the salt/water interaction.
    A second story Probably we have all been told that bacteria (usually) reproduce by, say, binary fission. This is more or less correct, and more or less the fact means that the rate of increase of bacteria at any time is directly proportional to the number of bacteria at that time. So twice as many bacteria "now" means that twice as many bacteria are being born now. This is certainly dreadfully simplified, but this approximation works in many circumstances. I wondered, when I first heard this fact, why, if, say, E. coli doubles rather rapidly, shouldn't the world be covered very soon by a layer of E. coli which is 40 feet thick? In fact,

    A single cell of the bacterium E. coli would, under ideal circumstances, divide every twenty minutes.
    (From Michael Crichton (1969) The Andromeda Strain, Dell, N.Y. p.247)
    But of course anything growing so rapidly in the real world (mold in a petri dish) enters a situation where the growth challenges the ability of the environment to support the thing. Most environments have a carrying capacity -- some sort of upper limit to the amount of the thing which can live in the environment. Differential equations can model this sort of situation fairly well, combined with the "exponential growth". We will see this.
    I went to a lecture today where someone who should know (a person who does both mathematics and biology) declared that the average human body contained 1014 bacteria. Indeed.

    What is a solution of a differential equation?

    Solving an example

    What is going on?

    Separable differential equations and another example

    Initial conditions and initial value problems

    A solution


    Is it useful?

    One example

    Another example
    dy/dx=2x with the initial condition (0,0): when x=0, then y=0.

    A last example (today!)
    dy/dx=xy2 with the initial condition y(0)=1.

    Monday, February 19 (Lecture #10)
    Exam coming
    And, yes, there will be a class on Thursday!

    Two silly (?) formulas
    The object of this lecture is to tell you about two formulas, one for arc length (section 8.1) and one for surface area (section 8.2). I called the formulas silly because of their limited usefulness, at least limited in the sense that "hand computation" is not very practical. Both arc length and surface area will be revisited in calc 3, where much better perspectives can be given for both.

    The philosophy behind the definite integral and its use
    Maybe the formulas are not totally silly. Both of them are illustrations of how definite integrals can be used to compute various quantities. The procedure (which we have already used in various area and volume situations, and also with work) represents an attempt to compute "something" complicated:

    1. Break up the complicated quantity into little pieces.
    2. Approximate the little pieces by something simple.
    3. Add up the little pieces, and take a limit.
    With some luck and skill and ... whatever, the errors which occur will be small, and as the number of subdivisions or pieces or whatever grow, the total error will get small.

    Arc length
    We're given a function, f(x), defined on the interval [a,b]. The quantity to be computed is the length of the graph, the curve y=f(x). This is called arc length. Here is the idea.
    Break up [a,b] into many little subintervals, whose length we will call dx (or delta x). "Above" each little subinterval is a little piece of the curve. The usual name for a little piece of curve is ds. If you magnify the little piece, as shown, well, the result is almost a right triangle. The curve length is still somewhat curvy, but, well, maybe I can approximate it by a straight line segment. The resulting picture is just about a right triangle. dy is the change in y (the function) when the input variable, x, Pythagoras then declares that (ds)2 should be the same as (dx)2+(dy)2. Therefore ds=sqrt{(dx)2+(dy)2). Let's rewrite what's inside the square root:
    So sqrt(=(dx)2(1+{dy/dx}2))=dx·sqrt(1+{f´(x)}2).

    Now we should add up these pieces and take limits. In this context, this is all done by writing a definite integral. So the arc length formula is absqrt(1+[f´(x)]2)dx. This is the official formula. Let's see how well it works with some examples.

    Line segment
    Maybe the simplest curve is a straight line segment. Let me "find" the length of the line segment joining (1,1) and (4,3). This should be the same as the distance from (1,1) to (4,3), which is (square root of the sum of the squares!) sqrt(13). Let's find this number using the calculus formula above.

    We need a formula for the line segment. The slope will be (3-1)/(4-1) which 2/3. So f(x)=(2/3)x+something. What will the "something" be? Since the line should pass through (1,1), when we put x=1, the result should be 1. Therefore (2/3)(1)+something=1, so something is 1/3. The formula is f(x)=(2/3)x+(1/3). The derivative is f´(x)=(2/3). Now the arc length is absqrt(1+[f´(x)]2)dx which is 14sqrt(1+[2/3]2)dx. The integrand is a constant, so the result is sqrt(1+[2/3]2)x]14=sqrt(1+[2/3]2)4-sqrt(1+[2/3]2)1=sqrt(1+[2/3]2)3. This is the same as sqrt(13).

    Maybe the next curve to look at is a circle, but we need the graph of a function so let's try to find the arc length of a semicircle. The semicircle I considered in class was the upper semicircle, radius 5, center at (0,0). For this curve, f(x)=sqrt(52-x2). Now I need sqrt(1+[f´(x)]2). So:
    f´(x)=(1/2)(52-x2)-1/22x using the Chain Rule. The 2's cancel, and we need to square the derivative, so:
    (f´(x))2=(52-x2)-1x2 but this is the same as

    to which we must add 1:
          x2     52-x2+x2      52
    1 + ----- = --------- = ------
        52-x2      52-x2     52-x2
    Finally we supposed to take the square root of this result, so that the integral we need to compute is -55 5/sqrt(52-x2)dx.

    This should look slightly familiar. The trig substitution x=5sin() makes this integral into 5d=5arcsin(x/5)+C. I am skipping the details because I've done many of these integrals already. Now evaluate the definite integral: 5arcsin(x/5)]-55=5arcsin(1)-5arcsin(-1), and (since I know arcsin(1)=Pi/2 and arcsin(-1)=-Pi/2) this works out to 5Pi, which is indeed half the circumference of a circle of radius 5.

    Problems in the book
    These two curves work out fairly well. But let's look at section 8.1, and some of the problems there. The problems mostly have the form, "Find the length of the graph of the function defined by the following formula" and here are some of the formulas:


    These are all very weird and ludicrous formulas. Let me try one, and maybe you will see some reason for the structure of the formulas.

    Doing a book problem
    Let's compute the length of f(x)=[x3/6]+[1/(2x)] as x goes from 1 to 3. We will need sqrt(1+[f´(x)]2). So:
    You should examine this computation with a bit of suspicion. The "middle" term has lots of coincidences. (The original formula of the function is designed for so these "coincidences" happen!) 2's cancel and x2's cancel. Now look:
    The darn original -1/2 has somehow changed to +1/2. And the -1/2 came from 2[x2/2](-[1/(2x2)]) so the +1/2 could be replaced by 2[x2/2](+[1/(2x2)]). A "miracle" has occurred! (Ehhh ... not very much of a miracle. It is really a designed algebraic event.) Therefore, let us replace the +1/2 by the stuff suggested and then see:
    1+(f´(x)}2)= [x4/4]+1/2+[1/(4x4)]=[x4/4]+2[x2/2](+[1/(2x2)])+[1/(4x4)]=the square of [x2/2]+[1/(2x2)]
    Therefore, sqrt(1+[f´(x)]2) becomes (take the square root of the thing that is squared): [x2/2]+[1/(2x2)].
    To find the arc length, integrate this from x=1 to x=3. The antiderivative is (relatively!) easy, and the result is [x3/3]-[1/(2x)]]13 etc. etc. (I'm really not too interested in the actual numbers here!)

    A parabola
    Let me find the arc length of y=x2 from x=0 to x=1. Here the arc length formula, absqrt(1+[f´(x)]2)dx becomes (since f(x)=x2 and f´(x)=2x) 01sqrt(1+4x2)dx. I can compute this, sigh, using a trig substitution. I could "try" 2x=tan() etc., etc. Indeed, I am advised by a friend that the antiderivative is

                               2 1/2
                     x (1 + 4 x )                           2 1/2
                     --------------- + 1/4 ln(2 x + (1 + 4 x )   )
    I think you know the friend I mean. Actually, since we are sophisticated now, if you are curious you can spot the ln(sec+tan) in that mess, etc. I don't feel like finishing. Let's try another example.

    I decided to save on pictures. The picture shown here will work just as well for the example below. They look sort of the same.

    Cubic curve
    Let me find the arc length of y=x3 from x=0 to x=1. Here the arc length formula, absqrt(1+[f´(x)]2)dx becomes (since f(x)=x3 and f´(x)=3x2) 01sqrt(1+9x4)dx. Now for the Hot News: there is no antiderivative of this function in terms of the standard functions. This actually can be proved. Therefore, I can't go any further "by hand". If I actually wanted to know the length, I would need to approximate the definite integral using some sort of numerical technique.

    The truth for arc length is that, more or less, the computability of the arc length integral using FTC is impossible almost all of the time! Therefore, from the elementary, student point of view, maybe this is all a waste of time. But, really, it isn't. As soon as you give me a definite integral and want to approximate the values, there are all sorts of strategies. So what's important is that arc length can be computed by a definite integral, and what's important for you to try to understand is the philosophy of going from the vague idea of arc length to the integral formula for the arc length. And that philosophy will now be displayed again as we get an integral formula for a certain type of surface area.

    Surface area
    Suppose we are again given a function y=f(x) defined on an interval [a,b]. I would like to "compute" (the quotes are because we will get a definite integral formula which will share the benefits and defects of the previous result) the surface area which results when the graph of y=f(x) is revolved around the x-axis.

    We will get our formula using the same philosophical approach. We can chop up [a,b] into many little pieces, each having length, say, dx. Then (the picture!) the little piece of arc length lieing over dx, which we called ds, will be revolved around the x-axis. This gets us a sort of ribbon. What is the area of that ribbon? We won't be able to compute it exactly, but maybe we can approximate the area of the ribbon nicely. Well, we can take the magic scissors (hey: I was able to draw the darn scissors almost correctly this time!) and cut the ribbon and then, sort of, almost, lay it out flat. The result will sort of, almost, be a rectangle. What are the dimensions of this rectangle? One side is the length of the piece of arc, ds. The other side is the circumference of a circle whose radius is f(x), the height of that part of the curve away from the x-axis. (The reason for the repeated "sort of, almost" is that this is actually a distortion of the true value - the ribbon really would not lie flat, and the ribbon really would not be more than an approximate rectangle. I will try later to address these sorts of slight (?) distortions.) So a piece of the surface area is 2Pi f(x) ds. We use a definite integral to get the total surface area and add everything up. The result for the area when the curve is revolved around the x-axis is img src="gifstuff/is11.gif" width=8>ab2Pi f(x)sqrt(1+[f´(x)])dx. Notice that sqrt(1+[f´(x)])dx (this uses what we had for ds).

    Here is a result from a long time ago: the surface area of a sphere of radius R is 4Pi R2. (This is the area of four "great circles" of the sphere, circles made by intersecting a plane with the center of the sphere.) I would like to verify this result using the surface area formula. I'll use the same semicircle as before: f(x)=sqrt(52-x2), with a=-5 and b=5. Please note that revolving this semicircle around the x-axis gets the area of the whole sphere of radius 5, so that the answer should be 4Pi(52).

    We need to compute ab2Pi f(x)sqrt(1+[f´(x)])dx. Notice that sqrt(1+[f´(x)])dx is what we called ds before, and we did compute ds in a previous example. We saw that ds was equal to 5/sqrt(52-x2)dx. But f(x)=sqrt(52-x2) so, wow! (yeah, wow) there is cancellation and the arclength becomes -55(2Pi)5dx which does indeed work out to 100Pi as it should.

    We can try to find the surface area which happens when y=x2 from x=0 to x=1 is revolved around the x-axis. So the formula ab2Pi f(x) sqrt(1+[f´(x)]2)dx becomes 012Pi x2 sqrt(1+4x2)dx and, oh my goodness! I can find an antiderivative of this. It is 2Pi multiplied by the following:

                     2 3/2             2 1/2
           x (1 + 4 x )      x (1 + 4 x )                            2 1/2
           --------------- - --------------- - 1/64 ln(2 x + (1 + 4 x )   )
                 16                32
    (Maybe you can tell where I got this from! If you wish, again the substitution 2x=tan() will "work") I don't feel like finishing.
    The picture supplied is sort of the same both for this example and for the next one. (I can't tell the difference too well!)

    Cubic curve
    We can try to find the surface area which happens when y=x3 from x=0 to x=1 is revolved around the x-axis. So the formula 012Pi f(x) sqrt(1+[f´(x)]2)dx becomes ab2Pi x3 sqrt(1+9x4)dx and, oh my goodness! I can find an antiderivative of this. If u=1+9x4 then du=36x3dx so (1/36)du=x3 and we have (2Pi/36) u1/2du=(2Pi/36)(2/3)u3/2+C= (2Pi/36)(2/3)(1+9x4)3/2+C. Then the surface area is (2Pi/36)(2/3)(1+9x4)3/2]01 and won't bother to finish.

    More "truth"
    x2 and x3 are two of only a few simple powers of x which give me integrands in the surface area formula that I can find antiderivatives of. (That's a horrible sentence!) If I want to compute surface areas for almost any "random" function defined by a formula, I'll need to use numerical approximations.

    I asked people to write an integral for the arc length of something like f(x)=5x7+2x as x goes from 1 to 3: just the integral, and do nothing with it.

    Maintained by and last modified 2/20/2007.